# Intuitive Understanding of the constant “$e$”

Potentially related-questions, shown before posting, didn't have anything like this, so I apologize in advance if this is a duplicate.

I know there are many ways of calculating (or should I say "ending up at") the constant e. How would you explain e concisely?

It's a rather beautiful number, but when friends have asked me "what is e?" I'm usually at a loss for words -- I always figured the math explains it, but I would really like to know how others conceptualize it, especially in common-language (say, "English").

related but not the same: Could you explain why $de^x/dx = e^x$ "intuitively"?

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Can you justify why this is not the same question? My answer to this question is the same as my answer to that question. –  Qiaochu Yuan Mar 9 '11 at 19:18
There are any number of "semi-natural" ways to arrive at $e$ (what you call "ending up at $e$"), such as compound interest, a function that is equal to its own rate of growth, etc. The compound interest one is particularly succint ("$e$ is the amount of interest you would have at the end of one year if you deposit one dollar, at 100% annual compound interest, compounded each and every instant."), though it may take some motivation to explain to the lay public (who has enough trouble grasping financial matters, it would seem), why the frequency of compounding matters. –  Arturo Magidin Mar 9 '11 at 19:25
@Qiaochu: IMHO the questions are not at all the same; they are complementary and so it is natural that they share the same basis for an answer. –  Eelvex Mar 9 '11 at 19:28
@Arturo Magidin: The number $e$ is what everybody wants to have (at least on his/her bankaccount). Imagine a world where the banks would pay interest continously... –  Fabian Mar 9 '11 at 19:29
@Fabian: Well, keep in mind that they would likely also charge interest continuously... –  Arturo Magidin Mar 9 '11 at 19:30

If someone asks me, "what is $e$?" I sketch the graph of $y=1/x$, draw a line segment from $(1,1)$ on the curve down to $(1,0)$ on the $x$-axis, and ask, how far to the right do I have to draw another vertical segment to rope off an area of 1? Anyone who is familiar with the idea of graphing a function can appreciate that definition, and it's not surprising that something with such a down-to-earth definition is going to turn up in lots of other places in Mathematics. And anyone who knows Calculus can be shown that all the other properties of $e$ and of $e^x$ and of $\log x$ can be derived from this one property of $e$.
For the somewhat-calculus-literate, your "related but not the same" question is what I'd go for: $e$ is the number for which the exponential function with that base is its own derivative.
Without calculus, I'd go for the notion of compound interest: With a rate $r$ per period, compounded $n$ times per period, $A$ grows to $A(1+\frac{r}{n})^n$ after 1 period; as $n\to\infty$, $A(1+\frac{r}{n})^n\to Ae^r$.
The interest method is exactly how I heard Sir Michael Atiyah explain it at a symposium. He explained it as investing $1 with an interest rate of 100%, both of which are easy numbers to wrap your head around. – JavaMan Mar 9 '11 at 19:31 It is my opinion there is no "intuitive understanding of the number$e$". Presumably, what you want to explain to your friends is not some mythical intuitive content of the number but some actual, concrete property it has which makes you appreciate it. Explain that. - Well put indeed! – Phonon Jul 29 '11 at 0:49 -1. I fully agree that there is no "intuitive understanding of a number," and that a more meaningful question would be, "How can I intuit the properties of the number$e$?" Now, we also seem to agree that the OP really meant to ask the latter question. So if that's the question, what's the answer? The response you gave does not provide one. I feel that this should be a comment rather than an answer. – Jesse Madnick Aug 5 '11 at 1:06 @JesseMadnick, I find the idea that one can intuit properties of a number as ununderstandable as that of reaching an intuitive understanding of the number$e$. – Mariano Suárez-Alvarez Apr 1 '13 at 3:33 @MarianoSuárez-Alvarez: Then what are the things that we can intuit, if not mathematical properties? – Jesse Madnick Apr 1 '13 at 4:07 I doubt I intuit any property of$e$whatsoever... what propertes of$e$do you intuit? (hopefully this verb does not exist...) – Mariano Suárez-Alvarez Apr 1 '13 at 4:14 Nobody has yet given a combinatorial way of visualizing$e,$so I thought I'd add one. The constant$\pi$is, of course, a ratio, that of the circumference of a circle to its diameter. If the corresponding ratios for regular polygons are used as a starting point, then$\pi$is defined by taking the limit as the number of sides of the polygon goes to infinity. In a similar way, the constant$e$is the limit of a ratio as a certain parameter is taken to infinity. Imagine a lottery in which, to enter, you must submit a sequence of$20$numbers in the range$1$to$20.$Order matters, and numbers can be used multiple times. So, for example, you can submit all$1\text{s}$if you like. The winning number could be any of$20^{20}$possibilities. Now imagine that a similar lottery has been set up in a different locale, except that, because of local superstition, use of the number$13$has been banned. Lottery entries still consist of$20$numbers in the range$1$to$20$, but only$19$number choices are available, giving$19^{20}$possible winning numbers. Clearly it's easier to win the second lottery, but by how much? The answer is the ratio$(1/19^{20})/(1/20^{20})=20^{20}/19^{20}\approx2.79.$You are$2.79$times as likely to win the second lottery as the first. To generalize, imagine comparing a lottery in which submissions consist of$n$numbers in the range$1$to$n$with a lottery in which one of the numbers in the range$1$to$n$is unlucky and may not be used. The probability of winning the second lottery is higher by a factor of$n^n/(n-1)^n.$The limiting value of this ratio as$n$goes to infinity is$e\approx2.718281828$. What does this have to do with rates of growth of exponential processes? Imagine a quantity that goes up by a fixed ratio every year, say$72\%.$Suppose we preferred to use months as units rather than years. It would be incorrect to simply divide the yearly rate by$12$to get$72\%/12=6\%$monthly growth, or, if we did so, we'd be describing a different process. The reason it's different is compounding. Growth of$6\%$per month means growth by a factor of$1.06$each month. In two months this translates to growth by a factor of$1.06\times1.06=1.1236$or an increase of$12.36\%,$which is bigger than$2\times6\%=12\%$since the$6\%$of quantity gained in the first month itself grows by$6\%$in the second month. Similarly, over the course of a year, the factor of increase is$1.06^{12}\approx2.0122,$which translates to$101.22\%$growth per year instead of$72\%.$More generally, if we prefer to use time units of$1/n$years, an increase of$72\%/n$every$1/n$years is bigger than an increase of$72\%$every year because of compounding. Specifically, it corresponds to growth by a factor of$(1+0.72/n)^n$every year, which, if you compute it for particular$n,$is bigger than$72\%$per year. Moreover, if we make the time unit smaller by making$n$bigger, the annual growth gets bigger because the compounding effect gets enhanced. In calculus we like talking about instantaneous rates of change, which involves make the time interval over which the change is measured arbitrarily small. If we do this in our problem, by taking the limit as$n$goes to infinity of$(1+0.72/n)^n,$we maximize the compounding effect, obtaining a growth factor of$2.05443,$or$105.443\%$per year. The limiting growth factor of$2.05443$turns out to equal to$e^{0.72}.$To relate this to the combinatorics problem, imagine a quantity that doubles every year, that is, that increases by$100\%$per year. If we instead split this$100\%$over$n$equal time intervals, we get a growth factor of$(1+1/n)^n=(n+1)^n/n^n.$As$n$goes to infinity, this has the same limiting value as$n^n/(n-1)^n,$namely$e.$Hence, because we are compounding on arbitrarily short time intervals, we grow by a factor of approximately$2.718281828$per year rather than by a factor of$2.$Interestingly, there's a related combinatorics problem in which$e$appears. Imagine, once again, a lottery in which$20$numbers in the range$1$to$20$must be chosen. But in this lottery all$20$numbers must be used. So our number selection boils down to choosing a permutation of the sequence$1,2,3,\ldots,20.$Now imagine a similar lottery with the extra stipulation that the first number may not be$1,$the second number may not be$2,$the third number may not be$3,$and so on. Such a permutation is called a derangement. Clearly there are fewer derangements than permutations, so the second lottery is easier to win. Once again, the probability of winning the second lottery is approximately$e$times bigger than the probability of winning the first. - Suppose you have a particle with the following property: starting at time zero (in seconds, s), the magnitude of its velocity$v = |\mathbf{v}|$(in meters per second, m/s) at time$t$is exactly equal to its distance$d$(in meters, m) from its starting place at the same time. At$1$m its speed is$1$m/s, at$2$m,$2$m/s and so on. What is the distance function$d = d(t)$of this particle as a function of time, you might ask? Since$v(t) = d^{\prime}(t) = d(t)$, then$d(t) = d_{0} \ e^{t}$, where$d_{0}$is some multiplicative constant. Since we haven't yet specified any initial data, we can (without loss of generality) simply take$d_0 = 1 \ \text{m}$and$v_{0} = 1 \ \text{m/s}$both at$t = 0 \ \text{s}$. The constant$e$is the magnitude of this particle's distance or velocity at time$t = 1$s. - it's unfortunate that you cannot hug someone or high five someone through a web browser yet, but I would at least like to say thanks for this wonderful answer :D – sova Aug 4 '11 at 17:06 When I was younger, I would often wonder about this while on road trips (except I would be driving toward my destination, not away from it). I since figured out what you have written here, but thanks for writing it! (+1) – The Chaz 2.0 Dec 6 '11 at 15:25 One way of understanding what is$e$, is to see it as a rate of growth. This article explains it very well. - what a wonderful link! many thanks for this. – sova Mar 9 '11 at 19:45 Thanks for the mention (I'm the author :)). You might want to check out betterexplained.com/articles/developing-your-intuition-for-math I cover 4 common definitions of e and how you can see the intuition behind them. – Kalid Mar 9 '11 at 21:47 Geometric interpretations help with the intuition, and I liked Gerry Myerson's explanation for that reason. Here's another geometric explanation you might give, using exponential decay. Start by imagining a process, such as radioactive decay, where at the end of every hour you have half the amount of material you started with at the begining of the hour. So if you start with 1 unit of material, then the amount of material remaining at hours 0, 1, 2, 3 is 1, 1/2, 1/4, 1/8. This process is described by$(1/2)^t$or$2^{-t}$. Or, you can imagine faster decay where you have only 1/3 of the material left at the end of each hour, so that the amounts are 1, 1/3, 1/9, 1/27 and the process is described by$(1/3)^t$or$3^{-t}$. You can then sketch these two functions; both asymptotically approach 0, with$3^{-t}$getting there faster than$2^{-t}$. You can then mention the amazing fact that the area bounded by either curve and the horizontal and vertical axes, although infinite in extent, has a definite finite area. This is highly plausible since the curve is approaching the horizontal so quickly. Tell them that if you calculate this area, you find that for$(1/2)^t$the area is bigger than 1, while for$(1/3)^t$, the area is smaller than 1. Then ask how can you adjust the decay rate, or equivalently, the fraction remaining after each hour, so that the area exactly equals 1? The answer turns out to be that$1/e\approx1/2.718$of the material should remain after each hour - a process described by$e^{-t}$. Not surprisingly, this is a number between 2 and 3. For the very curious and dedicated listener who knows about geometric series, you can justify the assertions that the areas are finite, and that the area is greater than 1 for$2^{-t}$and less than 1 for$3^{-t}$. For example, for$2^{-t}$you can get an overestimate of the area using rectangles:$1+1/2+1/4+\ldots=2$. So the area under the curve is finite. To show that for$2^{-t}$the area is bigger than 1, you can do an underestimate using rectangles:$1/2+1/4+1/8+\ldots=1$. To show that$3^{-t}$is smaller than 1, you can do an overestimate using trapezoids. If you break each trapezoid into a rectangle and a triangle, you get the overestimate $$\left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}\right)+\left(\frac{1}{9}+\frac{1}{2}\cdot\frac{2}{9}\right)+\ldots=2\left(\frac{1}{3}+\frac{1}{9}+\ldots\right)=1.$$ This provides some explanation for the magnitude of$e$. - thanks for a very cool answer, Will. Would it be more fair to say that the area under e^-t approaches 1, rather than is equal to it? – sova Aug 3 '11 at 2:21 @sova : That's a matter of definitions. I say that the proper integral$\int_0^b e^{-t}\,dt$approaches 1 as$b$approaches infinity or, in other words, that$\lim_{b\rightarrow\infty}\int_0^b e^{-t}\,dt=1$. I define "area under the curve" to be the value of the improper integral$\int_0^\infty e^{-t}\,dt$, which in turn is defined to be this limit, assuming it exists. – Will Orrick Aug 4 '11 at 2:21 I also believe this to be equivalent to the cited question about e^x, but will answer the explicit question for the sake of argument. To be concise, I remark to friends and students that e is the most important number in calculus, just as pi is the most important constant in geometry. Certainly these claims are arguable, but my friends/students don't argue! - Well, with that argument, you could also claim that$e$is home to a herd of unicorns... – Mariano Suárez-Alvarez Mar 9 '11 at 19:54 @mar With what argument?!? I usually don't justify the statement unless the listener has timefor something less-than concise. – The Chaz 2.0 Mar 9 '11 at 20:07 That is the argument I had in mind... the lack thereof. – Mariano Suárez-Alvarez Mar 9 '11 at 20:09 @MarianoSuárez-Alvarez - I think I get it now! Will leave the answer, for posterity's sake :) – The Chaz 2.0 Jun 10 at 13:49 Just to illustrate Gerry Myerson's answer. Yellow and red zones have same area. - It's easy to show that$\dfrac{d}{dx} 2^x = (2^x\cdot\text{constant})$. And$\left.\dfrac{d}{dx} 2^x\right|_{x=0} = \text{that constant}$. Since the graph of$y=2^x$gets steeper as$x$grows, the slope at$x=0$must be less than the slope of the secant line involving$x=0$and$x=1$. That latter slope is 1. Therefore the "constant" is less than 1. By thinking about$y=4^x$and considering the secant line involving$x=-1/2$and$x=0$, one sees that that "constant" is more than 1. Therefore 2 is too small, and 4 is too big, to be$e$. For$y=e^x$, the "constant" is exactly 1. (One can show that 3 is too big via the secant line at$x=-1/6$and$x=0$, but the arithmetic is a bit messy.) Similarly$2.5$is too small, via$x=0$and$x={}$. . . . I don't remember which number I used here. A positive number, obviously, and less than 1. Messy arithmetic again. - If$a > 0$, draw the curve$y = a^x$. You will notice it has a tangent line when it strikes the$y$-axis. If$a$is much larger than 1, the tangent line is steep. If$a$is small the slope is small. This slope depends continuously on$a$. The unique value making it 1 is$e$. This unlocks all of the magic if you think carefully. - Preface: The best first definition of e is one that nobody has yet given in this thread, namely, that it is the base such that the corresponding exponential function has a slope of unity at 0. If it is not immediately clear what this means, then sketch two or three exponential functions with different bases (say, 2, 3, and 1/2), note that they all pass through the point (0,1), and then note that they can be distinguished by what slope they have at that point (ie, what slope their tangent line has at that point). The “most convenient” one is the one that has a slope of unity. This base can be taken as the definition of e. However, once the edifice of one-variable calculus has been erected, it turns out to be convenient to define e in another manner, and this “first definition” is obtained as a theorem. Now, on to the answer: There is a joke, from the old West, about how to sell a covered wagon. As best as I recall, it goes something like this: “Say that the price is one hundred dollars. If the buyer doesn’t wince, add “just for the wheels”. If the buyer still doesn’t wince, add “for each one”…” You get the idea. In a similar vein, the answer to this question ought to be layered. The first best answer is that given by The Chaz (whose answer I have therefore up-voted), that is, just state that it is the most important number in calculus. Now, the audience mentioned by The Chaz is very limited, namely, his friends and students, however, the OP, naturally, wants us to consider the public at large, such as a casual acquantance encountered in an elevator who, stuck for something to say to fill the silence, asks you, whom they know to be an adept of mathematics, about this mysterious number e. The answer that The Chaz gives his friends and students is the best first answer for anyone, the best, as they say, elevator pitch (ie, something intelligible that you can say in the time that it takes an elevator to go between floors). If the questioner does not wince, that is, does not inquire further, then just leave it at that. If the questioner is not satisfied, then the next answer to give is the “exponential base whose tangent at 0 is 1” one (that I mentioned above). If the questioner does not inquire further, then just leave it at that. If the questioner is not satisfied, then the next answer to give is the compound-interst one. If the questioner does not wince, then leave it at that. If the questioner is still not satisfied, then give the calculus rate-of-change explanation – the velocity version by user02138 is especially good (and so I have up-voted it too). If the questioner is still not satisfied, then say, “You have reached the limit, no pun intended, of my anyone’s ability to explain this to you, short of you yourself learning calculus.” The questioner may then say, “I appreciate all you have said, but what I want to know is why e has the value it does, rathter than some other value.” You should then say, “That’s a very good question, and no one has ever answered that, but for that matter, no one has ever answered the corresponding question for π. For example, why is the third decimal digit of π 1, rather than, say, 2? But remember that mathematics is not unique in this regard. In Physics, there is a number, approximately equal to 1/137, the “intuitive understanding” of which is a genuine mystery. This number is called the fine structure constant. The big mystery is exactly what you are asking regarding e, namely, why it has the value that it does. There is a long-standing brouhaha about it, nothing like the placid acceptance of e in mathematics. After reading up on it, you just might come back to mathematics and look upon e as an old friend.” - I don't know why your post was downvoted. I think it's pretty witty! I voted it up. – Bruno Joyal Jul 28 '11 at 23:36 I had no strong feelings one way or the other about this answer until I read the final paragraph. Questions about mathematical constants like "Why is$\pi$what it is, rather than something else?" are not deep, they are trivial (or meaningless, depending on how you far you want to keep pushing on the "why" aspect), and I don't think they should be portrayed otherwise to the layperson, who won't know better. Thus, I have downvoted. – Zev Chonoles Jul 29 '11 at 0:24 I enjoyed this answer very much, and I'm sorry it was downvoted. Although I don't think I would ever be satisfied giving curt replies in an elevator to someone genuinely interested in Mathematics, you pointed out user02138's answer (the velocity one) to me -- [mind blown] meets [mental satisfaction] – sova Aug 4 '11 at 17:10 @Zev Chonoles: If it is true that the question of why mathematical constants are what they are is trivial, or even meaningless, then it seems like this metamathematical principle ought to appear somewhere as part of the ground rules. I’m willing to admit that there is at least some truth to this, but I think one could in good faith disagree. Would you also consider the question “Why is e irrational, rather than rational?” a trivial question? My guess is that many would not consider it trivial, but if it is not trivial, then can the question of its “choice” of digits be entirely trivial? – Mike Jones Aug 16 '11 at 4:47 Now, regarding your post here: it is not a "metamathematical principle" that needs to be built into the "ground rules", it's common sense. Choose your favorite definition of$e$; the reason$e$is what it is is that we have successfully made a precise definition, that refers to one and exactly one real number. I would say that it makes as much sense to wonder "what if$e$or$\pi$were rational?" as it does to wonder "what if$2=3$?". However, I think that the fact that$e$is irrational has interesting consequences, and that it is reasonable to wonder what those consequences are, – Zev Chonoles Aug 16 '11 at 5:45 There is a phenomenon called "exponential growth". We all have a feeling for it. Exponential growth of some quantity$Q$with time is characterized by the property that in equal time intervals the quantity$Q$increases by the same factor. Exponential growth can be slow or fast; therefore we have to have a "unit" in order to measure and compare the speed of growth in individual cases. Mathematicians tell us that the natural unit for this speed is encoded in a certain number$e\doteq 2.718$, in the same way that the unit of length was encoded in a rod of platinum and stored 1799 in a vault somewhere in Paris. - Professor Ghrist of University of Pennsylvania would say that e^x is the sum of the infinite series with k going from zero to infinity of (x^k)/k!. If you are interested in Euler's number then you should not miss his Calculus of a Single Variable Course on Coursera - The thing that is special is the exponential function$\mathrm{exp}$, which satisfies $$\mathrm{exp}(0)=1,\quad \mathrm{exp'}=\mathrm{exp}.$$ Then of course$e:=\mathrm{exp}(1)$, and because of$\mathrm{exp}(x+y)=\mathrm{exp}(x)\mathrm{exp}(y)$it makes sense to write$\exp(x)=e^x$. - The derivative of an exponential function is $$\left(a^x\right)'=\lim_{h\to0}\frac{a^{x+h}-a^x}h=\lim_{h\to0}\frac{a^h-1}ha^x=\ln(a)\ a^x,$$ where$\ln(a)$is some function of$a$. The number$e$is by definition the one such that$\ln(a)=1$, so that the derivative is the most "natural": $$\color{green}{\left(e^x\right)'=e^x}.$$ Informally, we can "invert" the limit and get: $$\ln(a)=\lim_{h\to0}\frac{a^h-1}h\to a=\lim_{h\to0}(1+h\ln(a))^{1/h}.$$ This has two useful consequences: 1.$\color{green}{e=\lim_{h\to0}(1+h)^{1/h}}$allows to compute$e$.$1.1^{10}=2.593\cdots\\1.01^{100}=2.7048\cdots\\1.001^{1000}=2.71692\cdots\\1.0001^{10000}=2.718145\cdots\\\cdots\\1.00000\cdots1^{100000\cdots0}=2.718281828459045235\cdots$1.$a=\lim_{h\to0}(1+h\ln(a))^{1/h}=\lim_{h'\to0}(1+h')^{\ln(a)/h'}=\lim_{h'\to0}\left((1+h')^{1/h'}\right)^{\ln(a)}=e^{\ln(a)}\$, and this shows that the logarithm function is the inverse of the exponential.